Density = (mass) divided by (volume)
We know the mass (2.5 g). We need to find the volume.
The penny is a very short cylinder.
The volume of a cylinder is (π · radius² · height).
The penny's radius is 1/2 of its diameter = 9.775 mm.
The 'height' of the cylinder is the penny's thickness = 1.55 mm.
Volume = (π) (9.775 mm)² (1.55 mm)
= (π) (95.55 mm²) (1.55 mm)
= (π) (148.1 mm³)
= 465.3 mm³
We know the volume now. So we could state the density of the penny,
but nobody will understand what we have. Here it is:
mass/volume = 2.5 g / 465.3 mm³ = 0.0054 g/mm³ .
Nobody every talks about density in units of ' gram/(millimeter)³ ' .
It's always ' gram / (centimeter)³ '.
So we have to convert our number for the volume.
(0.0054 g/mm³) x (10 mm / cm)³
= (0.0054 x 1,000) g/cm³
= 5.37 g/cm³ .
This isn't actually very close to what the US mint says for the density
of a penny, but it's in a much better ball park than 0.0054 was.
ANSWER and EXPLANATION
We want to identify if there will be an electric field and a magnetic field around the two sticks electrified by charges of opposite signs.
An electric field is a physical field that surrounds electrically charged particles and exerts a force on other charged particles in the surrounding.
This implies that the presence of electric charges on both sticks generates electric fields on them. Since the two charges are opposite, the electric force acting on them will be attractive.
Hence, there is an electric field.
A stationary charged object produces an electric field, as explained above, but will only produce a magnetic field if there is a motion of the object.
Hence, except the two sticks are caused to move, there will be no magnetic field around them.
Answer:
position 4
Explanation:
if the north is tilted away it gets no sunlight, so its winter
Because it just it dndbsnsnsbsnsnsnsnsnsnsnsnsn
Here since both children and merry go round is our system and there is no torque acting on this system
So we will use angular momentum conservation in this
now here we have
Now when children come to the position of half radius
then we will have
now from above equation we have
